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In an isosceles triangle AB = AC = 25 cm BC = 14 cm. What is the length of altitude from A on BC?
Here altitude from A to BC is AD, we know in isosceles altitude on nonequal sides also median,
so, BD = CD = BC/2 = 14/2 = 7cm
Now, we applying Pythagoras theorem in △ABD,
AB^{2} = BD^{2} + AD^{2}
(25)^{2} = (7)^{2} + AD^{2}
AD^{2} = 625 − 49
AD^{2} = 576
AD = 24cm.
The area of two equilateral triangle are in the ratio 196 : 169. What is the ratio between their perimeters?
Ratio of perimeters = ratio of sides
D and E are points on the sides AB and AC respectively of a DABC such that DE  BC find value of x if AD = (7x  4) cm, AE = (5x  2) cm, DB = (3x  4) cm, EC = 3x cm
∵ DE ║ BE
∴ DADE ~ DABC
⇒
⇒ adjusting this expression, we have
⇒
⇒
⇒ 21x^{2}  12x = 15x^{2} − 6x − 8 + 20 x
⇒ 6x^{2 }− 26x + 8 = 0
⇒ 6x^{2} −24x − 2x + 8 = 0
⇒ 6x(x − 4) − 2(x − 4) = 0
When,
∴
In the given figure in DPQR, ST  QR, so that PS = (7x  4) cm, PT = (5x  2) cm, QS = (3x + 4) cm, RT = 3x cm.
What is the value of x?
In ΔPQR and ΔPST
⇒ According to Thales’ theorem
Subtracting unity from both sides, we have,
⇒ 21x^{2 } 12x = 15x^{2}  6x + 20x  8
⇒ 6x^{2}  26x + 8 = 0
⇒ 3x^{2}  13x + 4 = 0
⇒ 3x^{2}  12x  x + 4 = 0
⇒ 3x(x  4)  1(x  4) = 0
⇒ (x  4) (3x  1) = 0
∴ x = 4
A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. What is the width of the street?
In ΔABE,
AE^{2} + AB^{2} = BE^{2}
⇒ 9^{2} + AB^{2} = 15^{2}
⇒ AB^{2} = 225  81
= 144
⇒ AB = 12 m
Similarly,
In DBCD
⇒ BC^{2} = BD^{2}  DC^{2} = 15^{2}  12^{2} = 92
BC = 9m
∴ width of street = (AB + BC) = 12 + 9 = 21 m
The corresponding altitude of two similar triangle are 6 cm and 9 cm respectively. What is ratio of their areas?
Ratio of areas = (ratio of altitudes)^{2}
In the given figure, ΔODC ~ ΔOBA, ∠BOC = 115°, ∠CDO = 70°. What is the value of ∠OAB?
∠ CDO + ∠DCO = 115°
⇒ ∠DCO = 115°  70° = 45°
∠OAB = 45° (∴ DODC ~ DOBA)
A tree is broken of 6 m from the ground and its top touches the ground at a distance of 8 m from the base of the tree. What is the original height of the tree?
A is the point of cleavage C is the point where top of three touches the ground
In ΔABC
AB^{2} + BC^{2} = AC^{2} (Pythagoras’ Theorem)
6^{2 }+ 8^{2} = AC^{2}
= AC = 10 m
∴ Total (original) height = 10 m + 6 m = 16 m
An exterior angle of a triangle measures 110° and its interior opposite angles are in the ratio 2 : 3. Which is the largest angle of the triangle?
Let the angles be 2x and 3x respectively
∴ 2x + 3x = 110°
⇒ 5x = 110°
⇒ x = 22°
∴ Angles are 44°, 66°, and 70°
What is the perimeter of a rhombus the length of whose diagonal are 16 cm and 30 cm?
Let the side of rhombus be x cm
Now, we know that the diagonals of rhombus bisect each other at right angles.
∴ By using Pythagoras’ Theorem
= 64 + 225 = 289
⇒ x = 17 cm
∴ Perimeter = 4 × x = 4 × 17 = 68 cm.
In the given figure BC is parallel to DE. Area of DABC is 25 cm^{2}. Area of trapezium BCED is 24 cm^{2}, if DE = 14 cm
What is the length of BC?
∵ BC ║ DE
∴
[Ar(DADE) = Ar(DABC) + Ar(¨BCED)
⇒ BC = 10 cm
A man goes 10 m due south and then 24 m due west. How far is he from the starting point?
In ΔABC
AC^{2} = AB^{2} + BC^{2} (Pythagoras’ theorem)
⇒ AC =
A vertical pole 12 m long casts a shadow of 8 m long on the ground. At the same time, a tower casts the shadow 40 m long on the ground. What is the height of the tower?
Both Ds are right angled, and also the elevation of coming sunlight will be same.
∴ All the angles of the ΔABC and ΔPQR will be correspondingly identical.
∴ ΔABC ~ ΔPQR
⇒ AB/PQ = BC/QR
⇒ 12/PQ = 8/40
⇒ PQ = 12 × 5 = 60m
In an isosceles triangle ΔABC if AB = AC = 13 cm and altitude from A on BC is 5 cm. What is the length of BC?
In ΔABD
AB^{2} = BD^{2} + AD^{2}
⇒ BD =
= 12 cm
∵ D bisects BC
∴ BC = 2 × BD
= 2 × 12 = 24 cm
The Perimeter of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm. Then what is the corresponding side of the second circle?
We know that if, ΔABC ~ ΔPQR, then
or, Applying componendo and dividendo, it can be clearly seen that,
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